can someone help me with locus (1 Viewer)

[wolf7]

can some one tell me how this locus problem is, i have trouble understanding it

find the equation of the parabola with coordinates of the vertex (0,0) and equation of the axis x=0 and passing throught the point (-8,2)

also

find the equation of the parabola with coorinates of teh vertex (0,0) and equation of the axis x = 0 and passing throught the point (-1,7)

 

[Riviet]

Heh, took a few minutes to figure out what the question is on about lol.
Basically I think when it says "equation of the axis x=0" it means the parabola is symmetrical by the line x=0, implying that the curve is either concave up or concave down with vertex at the origin since these are the only two possibilites that have axis of symmetry of the x=0 line.

Now for the first question, it says the parabola passes through (-8,2) and this is in the 2nd quadrant, therefore, it must be concave up to pass through this point.

Similarly for the second question, the parabola passes through (-1,7), which is also in the 2nd quadrant, hence it must be concave up to pass through this point.

Hint to help you finish off: Since both parabolas pass through (0,0) they both must have equation ax²+bx, where c=0.

You didn't ask for full solution, only an explanation of question, so you can do the rest.

 

[word.]

... where b = 0 as well.
y = ax² + bx = x(ax + b) which implies 2 solutions to the eqn.

the vertex is at (0,0) so it is a parabola in the form y = ax²
so, basically just sub the point (x,y) in and solve for a

 

[sasquatch]

To correct word, the equation of a parabola is in the form,

(x - h)^2 = 4a(y - k)

Where (h,k) is the vertex and a is the focal length. The part of the question saying the equation of the axis x = 0 is pretty useless as from the position of the vertex (0,0) it is becomes pretty obvious.

The points they give you, aid you in finding the value of a (the focal length). If you still dont understand this, or wish to see the solution, just reply.

 

[word.]

...
listen my genius Einstein friend; you don't need the focal length whatsoever.
my 'a' was just the constant to be found which simplifies things.

let me simplify this for you: the parabola is in the form y = kx² where k is a constant.
k in this case would be equal to 1/(4a) where a is the focal length... but it doesn't matter

and the part where it says the axis is at x = 0 does matter because you can have sideways parabolas in the form x = ky² or (y - k)² = +/- 4a(x - h) with vertex (h,k) and focal length a. since h = 0, k = 0 is possible, the removal of

...and equation of the axis x=0...

will provide an extra solution,
namely x + 2y² = 0 (axis: y = 0)

 

[acmilan]

word. is quite right, why complicate things with information you dont need.

 

[Riviet]

Exactly. :uhhuh:

 

[sasquatch]

Ok your right bout the axis bit, but how does using (x - h)^2 = 4a(y - k) complicate things?

Using y = ax^2 + bx + c means you need to find three coefficients: a and b, but using the equation (x-h)^2 = 4a(y-k) you only need to find y. Also using your method it would be a very complciated task finding the equation of a parabola at a vertex of say: (10, -6). And also most locus problems based on parabola's arent as simple as, vertex (0,0), axis: x = 0, so by having a method that will apply in any situation would be the best thing.

 

[insert-username]

Ok your right bout the axis bit, but how does using (x - h)^2 = 4a(y - k) complicate things?

Because if your vertex is at the origin, which often it is, all you need is x² = 4ay or y² = 4ax. If your parabola has a different vertex, then by all means apply the other formula - it's meant to be used for parabolas with vertexes not at the origin. But if the vertex is at the origin it's faster to leave out the h and the k.


I_F

 

[sasquatch]

ah man thats obvious...............................

:|

(x - h)^2 = 4a(y - k) where (h,k) is the vertex.....

if the vertex is (0,0)

then unless your an idiot, its obvious that x- h = x and y - k = y....

so i dont know what your trying to say..

 

[acmilan]

ah man thats obvious...............................

:|

(x - h)^2 = 4a(y - k) where (h,k) is the vertex.....

if the vertex is (0,0)

then unless your an idiot, its obvious that x- h = x and y - k = y....

so i dont know what your trying to say..

There would have been nothing wrong with your post if you hadn't said:

"To correct word. ..." and "The part of the question saying the equation of the axis x = 0 is pretty useless"

"To correct word. ..." implies he did something wrong, which he didnt. In fact he provided a simpler method.

"The part of the question saying the equation of the axis x = 0 is pretty useless" is wrong for obvious reasons

 

[sasquatch]

yes yes i appologised for the axis thing....

when i said to correct word i was just giving a general formula that could be used in any situation....

my main point is though..regarding locus and parabolas, your generally asked much more harder questions that that, i.e. finding the equation of a parabola with a focal length of something, or finding the equation of a parabola with a shifted vertex. Using words method, this would be a very complicated task as the formula: y = ax^2 + bx + c, does not indicate where the vertex is, you need to rely on x = -b/2a and then subbing the point of x on the axis of symmetry to find the y-coordinate, hence working backwards is pretty hard. The equation (x - h)^2 = 4a(y - k) provides a VERY simple method for finding the equation of a parabola.

Using words method you would need to:

- Determine weather or not the parabola cuts the y-axis (for c)
- Find the value of a and b, by using x = -b/2a, if the parabola was not symetrical about x = 0, then you would have to rely on simulateous equations to find the values of a and b, which would also involve substituing the coordinates of a point it passes into x and y.
- Also if the parabola was shifted, it would be hard to find a value for c.

Using the method i described.

- Substitute the coordinates of the vertex into (h,k)
- Substitute the cordinates of a point the parabola passes to solve for a.

Which seems easier?

 

[insert-username]

sasquatch, people do things in different ways. Sometimes it's simpler to express a parabola in general form rather than 4ay form. Yes, when you're given a vertex, you would probably be expected to use 4ay form, but general form works just as well - especially if you're not given a vertex or you have to derive the function. In that situation, your method can't be used unless you find the vertex or rearrange the equation, so it can't be "used in any situation". Use whichever fits the question better and whichever you're better at. Neither method is in any way wrong. And:


- Determine weather or not the parabola cuts the y-axis (for c)

- Also if the parabola was shifted, it would be hard to find a value for c.

- Find the value of a and b, by using x = -b/2a, if the parabola was not symetrical about x = 0, then you would have to rely on simulateous equations to find the values of a and b, which would also involve substituing the coordinates of a point it passes into x and y.


The question tells you the vertex is at the origin, therefore c = 0 and the y-intercept is at (0,0), like word said.


I_F

 

[Riviet]

Use whichever fits the question better

Exactly, it can be referred to as common sense.

 

[sf_diegoxrock]

is the answer x^2 = 32y ?

 

[sf_diegoxrock]

and x^2 =1/7 y??

i used x^2 = 4ay its not complicated at all